Absolute direction in space Rotation, from my understanding, is basically the "exchanging of different spatial dimensions with eachother", with $x^2+y^2=d^2$ being the "relationship" between any two spatial dimensions, aka. if your point that you are rotating is initially at $[x,y]$ relative to the centre of rotation, then $x$ and $y$ get "exchanged" in a rotation while having the sum of their squares be constant/remain unchanged. If you replace the exponents with $1$ (aka. no exponent), there seems to be a fixed/absolute rotation/orientation of those two spatial dimensions such that they are detectable by entities living in that space, same with 3. My question is why is it that a non-absolute orientation of spatial dimensions seems to only apply to the exponent 2?
 A: I actually think that this is a meaningful question, but that your focus is off.
So yes: rotations are those linear transforms $$x \mapsto \bar x = a x + b y\\y\mapsto \bar y = c x + d y$$ which preserves the metric $\sqrt{x^2 + y^2} = \sqrt{\bar x^2 + \bar y^2}.$ More specifically, we can always choose $a = d = \cos\theta$ with $b = -c = \sin\theta$ to realize these by the trigonometric version of the Pythagorean theorem, $\cos^2\theta + \sin^\theta = 1.$
Now, $d(a, b) = (a_x - b_x) + (a_y - b_y)$ is not a metric (please read the linked article above!) but there is a metric which is like it, the Manhattan metric, based on "counting city blocks" if you cannot move diagonally: $d(a,b) = |a_x - b_x| + |a_y - b_y|.$ And yes, the only linear transforms which preserve the metric (possibly the only transforms which preserve the metric!) are translation (continuous), reflections, and rotations by 90 or 180 degrees. These transforms are called "isometries". So that might seem very strange: is the Euclidean metric the only one which has a continuous family of isometries?
The answer is no, and in fact its one-dimensional continuum of isometries in $\mathbb R^2$ pales in comparison to the 4-dimensional continuum of linear isometries of another metric. 
This metric is called the trivial metric, $d(a, b) = 0\text{ if } a = b \text{ else } 1.$ It is "trivial" because the "distance" it measures is just whether things are different, but it doesn't measure how different they are in any appreciable sense. But it still satisfies all of the metric axioms. In this case we only have to map the point $(1, 0)$ to $(a, b)$ and the point $(0, 1)$ to some $(c, d)$ [just as above!] but with $(a, b)$ and $(c, d)$ not being on a line that intersects with $(0, 0)$: $ad \ne bc$. 
Now: we do not know why Nature has equipped us with the particular metric we have, which puts us on a $3+1$ Lorentzian manifold equipped with the Lorentz metric $\operatorname{diag}(1, -1, -1, -1).$ There are a couple hints which may or may not pan out into bigger questions, like the fact that the Lorentz group of transforms on our 3+1 geometry, a group named $\operatorname{SO}^+(1, 3),$ ("restricted special orthogonal group $1,3$") turns out to be the exact same group as a different group named $\operatorname{PSL}(2, \mathbb C),$ ("projective special linear group in 2 complex dimensions" or more simply "the Möbius group"). This sort of idea, that Nature fundamentally is built out of complex numbers (or maybe quaternions) and that spacetime is somehow built out of how they connect together, underlies a bunch of somewhat off-mainstream work in quantum gravity, in particular some parts of loop quantum gravity and all of twistor theory. The reason that the Euclidean metric pops up is that Nature seems to have given us a Lorentzian (hence pseudo-Riemannian) manifold which, if we only consider distances of two points at the same time, works out as the Euclidean distance. Maybe there's a more fundamental reason why that's happened; nobody knows.
But of course in any question like this, even if we can answer that question of "why SO(1, 3)?" we will then have the question of, "well, why that?" so it doesn't give practicing physicists much loss of sleep.
A: I suppose what you say is true, but the observer has to agree with you on the orientation of your axes.  That orientation is arbitrary, and is not fixed in any way to an absolute direction in space.
It's pretty much the same as saying:  If I'm standing next to you on the ground I can watch you turn yourself around.  I still can't say anything about the absolute direction in space you are facing.
A: Values that do not change after a transformation such as rotation are called Lorentz scalars. While the components of vectors and tensors will vary under such a transformation, the value of a scalar is something everyone must agree on under these transformations.
In the case you give, the length of a rod is $d$ whether it is perpendicular, parallel or some combination therein to you. The length of the rod is given by the dot product of the coordinates used to define the position of the rod:
$$
d=\sqrt{\mathbf{x}\cdot\mathbf{x}}
$$
If we rotate this by some arbitrary angle, then we should very well expect
$$
d=\sqrt{\mathbf{x}'\cdot\mathbf{x}'}
$$
where $\mathbf{x}'=\Lambda \mathbf{x}$ represents the rotation transformation of this vector. This puts a constraint on $\Lambda$ such that multiplying $\Lambda$ and its inverse, $\Lambda^{-1}$, we get 1 back (i.e., $\Lambda^{-1}\mathbf{x}'=\Lambda^{-1}\Lambda\mathbf{x}=\mathbf{x}$).
The only way to get a scalar from a vector is through the inner product:
$$
d^2=\mathbf{x}\cdot\mathbf{x}=x^2+y^2
$$
(to restrict to 2D). We can also get powers of this, but that'd give us things like $\propto x^{2n}$ for the $n$th power, so we cannot get $\propto x^3$ or $\propto x$ from this, so the $\propto x^2$ scalar is special in that it is the simplest scalar that is preserved under rotations.
